# Simple derivative

• Jun 26th 2013, 11:54 PM
hacker804
Simple derivative
Hi I need to find the derivative of $\displaystyle \frac{1}{\sqrt{1-x^2}}$.I found it as:
$\displaystyle \frac{d}{dx} (\frac{1}{\sqrt{x^{2}-1}})\newline$
$\displaystyle \frac{d}{dx} (x^{2}-1)^{-\frac{1}{2}}=-\frac{1}{2}(x^{2}-1)^{-\frac{1}{2}-1} \newline$
$\displaystyle =-\frac{1}{2(x^2-1)^\frac{3}{2}}$

But the answer by the book is $\displaystyle -\frac{x}{(x^2-1)^\frac{3}{2}}$

Help.
• Jun 27th 2013, 12:57 AM
Prove It
Re: Simple derivative
Your function is \displaystyle \displaystyle \begin{align*} y = \left( x^2 - 1 \right) ^{-\frac{1}{2}} \end{align*}, which is a COMPOSITION of functions. So to differentiate this you will need to use the chain rule.

You have done the outer derivative correctly, but you have forgotten that you also need to find the inner derivative, which is \displaystyle \displaystyle \begin{align*} \frac{d}{dx} \left( x^2 - 1 \right) = 2x \end{align*}. Multiply this with the result you got.
• Jun 27th 2013, 01:13 AM
hacker804
Re: Simple derivative
Well I have not studied chain rule yet.thanks for the help.
• Jun 27th 2013, 08:37 AM
hollywood
Re: Simple derivative
I'd be interested to know how to do this problem without the chain rule. Does anybody have any ideas?

- Hollywood
• Jun 27th 2013, 09:44 AM
HallsofIvy
Re: Simple derivative
I can't see why you would want not to use the chain rule.

By the way, did you notice that your original function was $\displaystyle \frac{1}{\sqrt{1- x^2}}$ while the function you tried to integrate was $\displaystyle \frac{1}{\sqrt{x^2- 1}}$. Those are not at all the same thing!
• Jun 27th 2013, 06:50 PM
hollywood
Re: Simple derivative
He said he hadn't studied the chain rule yet, and I couldn't see a non-chain-rule solution. So I was curious.

I didn't notice the change in the function.

- Hollywood
• Jun 28th 2013, 11:27 AM
johng
Re: Simple derivative
Hi,
Since you haven't studied the chain rule yet, I assume you're just starting derivatives. So perhaps you want to find the derivative by definition? The attachment shows how to do this.

Attachment 28690